Question: The base of a solid $S$ is the region bounded by the ellipse $4x^2+9y^2=36$. $y$ $x$ $(3,0)$ $(0,-2)$ ${4x^2+9y^2=36}$ Cross-sections perpendicular to the $y$ -axis are semicircles. Determine the exact volume of solid $S$.
Let's graph the base of the solid. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $2x$. $y$ $x$ $2x$ $(3,0)$ $(0,-2)$ $(x,y)$ $(-x,y)$ ${4x^2+9y^2=36}$ Since each cross-section is perpendicular to the $y$ -axis, the independent variable is $y$. If $A$ denotes the area of each cross-section as a function of $y$, the volume $V$ of solid $S$ is $ V=\int_a^b A(y) \,dy$. To determine the area $A$ as a function of $y$, first express $A$ in terms of $x$. Since the semi-circular cross-section rests on the rectangle pictured above, the diameter of the semi-circle is $2x$. The radius of the semi-circle is $x$. $x$ $2x$ The area $A$ of the semi-circle is $A=\dfrac12\cdot\pi x^2=\dfrac\pi2 x^2$. What is $A$ as a function of $y$ ? The corner point $(x,y)$ of the rectangle lies on the ellipse $4x^2+9y^2=36$. Let's rewrite the equation as $ x^2=\dfrac14\left(36-9y^2\right)=\dfrac94\left(4-y^2\right)$. Now we can express $A=\dfrac\pi2x^2$ in terms of $y$ as $A(y)=\dfrac{9\pi}8\left(4-y^2\right)$. Can you express the volume $V$ of solid $S$ as a definite integral? Since $y$ goes from $-2$ to $2$, the volume formula $ V=\int_a^b A(y) \,dy$ gives us the definite integral $\begin{aligned} V&=\int_{-2}^2 \dfrac{9\pi}8\left(4-y^2\right) dy \\\\ &=\dfrac{9\pi}8\int_{-2}^2 \left(4-y^2\right) dy \end{aligned}$ Since we're integrating an even function over a symmetric interval, we can rewrite the integral as $ V=\dfrac{9\pi}4\int_0^2 \left(4-y^2\right) dy$. What is the value of the integral? $\begin{aligned} V&=\dfrac{9\pi}4\int_0^2 \left(4-y^2\right) dy \\\\ &=\dfrac{9\pi}4\left[4y-\dfrac13y^3\right]_0^2 \\\\ &=\dfrac{9\pi}4\left[4(2)-\dfrac13(2)^3-\left(4(0)-\dfrac13(0)^3\right)\right] \\\\ &=\dfrac{9\pi}4\left[8-\dfrac83\right] \\\\ &=12\pi \end{aligned}$